poj 2159 Ancient Cipher 简单的密码转化问题

Ancient Cipher

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 22033 Accepted: 7472

Description

Ancient Roman empire had a strong government system with various departments, including a secret service department. Important documents were sent between provinces and the capital in encrypted form to prevent eavesdropping. The most popular ciphers in those times were so called substitution cipher and permutation cipher. 
Substitution cipher changes all occurrences of each letter to some other letter. Substitutes for all letters must be different. For some letters substitute letter may coincide with the original letter. For example, applying substitution cipher that changes all letters from 'A' to 'Y' to the next ones in the alphabet, and changes 'Z' to 'A', to the message "VICTORIOUS" one gets the message "WJDUPSJPVT". 
Permutation cipher applies some permutation to the letters of the message. For example, applying the permutation <2, 1, 5, 4, 3, 7, 6, 10, 9, 8> to the message "VICTORIOUS" one gets the message "IVOTCIRSUO". 
It was quickly noticed that being applied separately, both substitution cipher and permutation cipher were rather weak. But when being combined, they were strong enough for those times. Thus, the most important messages were first encrypted using substitution cipher, and then the result was encrypted using permutation cipher. Encrypting the message "VICTORIOUS" with the combination of the ciphers described above one gets the message "JWPUDJSTVP". 
Archeologists have recently found the message engraved on a stone plate. At the first glance it seemed completely meaningless, so it was suggested that the message was encrypted with some substitution and permutation ciphers. They have conjectured the possible text of the original message that was encrypted, and now they want to check their conjecture. They need a computer program to do it, so you have to write one.

Input

Input contains two lines. The first line contains the message engraved on the plate. Before encrypting, all spaces and punctuation marks were removed, so the encrypted message contains only capital letters of the English alphabet. The second line contains the original message that is conjectured to be encrypted in the message on the first line. It also contains only capital letters of the English alphabet. 
The lengths of both lines of the input are equal and do not exceed 100.

Output

Output "YES" if the message on the first line of the input file could be the result of encrypting the message on the second line, or "NO" in the other case.

Sample Input

JWPUDJSTVPVICTORIOUS

Sample Output

YES

Source

Northeastern Europe 2004

题目给的意思及其不明确~~最后wa了好几次才经同学提醒才过的~~~  后来发现 文中画红线的是关键字  原文的意思是这样的，文字加密技术有常见的两种形式

1   substitution cipher

2   permutation cipher

第一种形式的字母的替换，将字母向后移几位，变成其它的字母，这样来进行求解。 但是本道题中每个字母向后移位的个数互不相同。

第二种形式的字母的互换，字母之间会相互交换位置。

算法 其实 这两种形式合到一块就是统计每个字母出现的个数，再将这些个数排序，对应的字母的个数能相互匹配就可以。

例如

AC  其中A --1个   C--1个

BM 其中B--1个   M--1个   个数一一对应 成立

 

AA 其中A--2个 

BC其中B--1个  C--1个  个数没有一一对应 不成立

post code：

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;char a[100],b[100];int  a1[100],b1[100];int main(){    while(gets(a))    {       gets(b);         memset(a1,0,sizeof(a1));       memset(b1,0,sizeof(b1));       int len=strlen(a),i;       for(i=0;i<len;i++)   //统计每个字母的个数       {          a1[ a[i]-'A' ]++;          b1[ b[i]-'A' ]++;                          }       sort(a1,a1+26);     //将个数进行排序       sort(b1,b1+26);       for(i=0;i<26;i++)     //比较个数       {        if(a1[i]!=b1[i])break;                       }       if(i==26)printf("YES\n");         else printf("NO\n");                      }    }